Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
typedef long long lint;
typedef unsigned int uint;
typedef pair<int, int> pii;
typedef pair<lint, lint> pll;
typedef unsigned long long ulint;
#define endl '\n'
#define fst first
#define sed second
#define pb push_back
#define mp make_pair
#define rint register int
#define newline putchar('\n')
#define leave_space putchar(' ')
#define all(x) (x).begin(), (x).end()
#define reveal(x) cerr << #x << " = " << (x) << endl
#define rep(it, f, e) for (rint it = (f); it <= (e); ++it)
#define per(it, f, e) for (rint it = (f); it >= (e); --it)
const lint INF = 1e18;
const int MAXN = 1e5 + 5;
int a[MAXN], b[MAXN], n, k;
struct Segment_tree {
	#define lson tree[root].ls
	#define rson tree[root].rs
	#define stdl (tree[root].sl)
	#define stdr (tree[root].sr)
	#define mid ((stdl + stdr) >> 1)
	struct Node {
		int sl, sr, ls, rs;
		lint val, rval, cover;
		Node () : ls(-1), rs(-1), cover(-1) {} 
	};
	vector<Node> tree;
	inline int new_Node(int l, int r) {
		static int root = 0;
		tree.pb(Node());
		stdl = l, stdr = r;
		tree[root].val = 0;
		tree[root].rval = l;
		return root++;
	}
	inline void update(int root) {
		if (~tree[root].cover) {
			tree[root].val = tree[root].cover;
			tree[root].rval = stdl + tree[root].cover;
		} else {
			tree[root].val = min(tree[lson].val, tree[rson].val);
			tree[root].rval = min(tree[lson].rval, tree[rson].rval);
		}
	}
	inline void pushdown(int root) {
		int L = lson, R = rson;
		if (lson == -1) L = new_Node(stdl, mid);
		if (rson == -1) R = new_Node(mid + 1, stdr);
		lson = L, rson = R;
		/*
			lson = new_Node(stdl, mid); ? RE
		*/
		if (~tree[root].cover) {
			tree[lson].cover = tree[rson].cover = tree[root].cover;
			update(lson), update(rson), tree[root].cover = -1;
		}
	}
	inline void replace(int root, int l, int r, lint p) {
		if (l <= stdl && stdr <= r) {
			tree[root].cover = p;
			update(root);
			return;
		}
		pushdown(root);
		if (l <= mid) replace(lson, l, r, p);
		if (r >= mid + 1) replace(rson, l, r, p);
		update(root);
	}
	inline lint query(int root, int l, int r) {
		if (l <= stdl && stdr <= r) {
			return tree[root].rval;
		}
		pushdown(root);
		lint ret = INF;
		if (l <= mid) ret = min(ret, query(lson, l, r));
		if (r >= mid + 1) ret = min(ret, query(rson, l, r));
		return ret;
	}
	#undef lson
	#undef rson
	#undef stdl
	#undef stdr
	#undef mid
	/*lint tree[10086];
	inline void new_Node(int l, int r) {
		rep (i, l, r) {
			tree[i] = i;
		}
	}
	inline void replace(int root, int l, int r, lint p) {
		rep (i, l, r) { tree[i] = i + p; }
	}
	inline lint query(int root, int l, int r) {
		lint ret = INF;
		rep (i, l, r) { ret = min(ret, tree[i]); }
		return ret;
	}
	inline lint query2(int root, int l, int r) {
		lint ret = INF;
		rep (i, l, r) { ret = min(ret, tree[i] - i); }
		return ret;
	}*/
}	Tree;
inline lint getquery(lint T) {
	return min(Tree.query(0, 0, T) + k - T, Tree.query(0, T, k - 1) - T);
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(NULL), cout.tie(NULL);
	cin >> n >> k;
	Tree.new_Node(0, k - 1);
	lint ans = 0, res, T = 0;
	rep (i, 1, n) {
		cin >> a[i] >> b[i];
		ans += (a[i] <<= 1);
	}
	rep (i, 1, n) {
		if (b[i] == 1) {
			if (a[i] > k) {
				puts("-1"); return 0;
			}
			lint l = (T + 1) % k;
			lint r = (T + a[i] - 1) % k;
			lint val = getquery(T);
			if (l > r) {
				Tree.replace(0, 0, r, INF);
				Tree.replace(0, l, k - 1, INF);
			} else {
				Tree.replace(0, l, r, INF);
			}
			Tree.replace(0, T, T, val);
			T = (T + a[i]) % k;
		} else {
			T = (T + a[i]) % k;
			Tree.replace(0, T, T, getquery(T));
		}
	}
	res = Tree.tree[0].val;
	cout << (res < INF ? res + ans : -1) << endl;
	return 0;
}